Stuck on a simple piecewise linear element problem

Question

See the schematic of interest on a picture. \$V_S = 5 \,\rm{V}\$, \$R_S = 4700 \,\Omega\$, \$R_P = 5800 \,\Omega\$. The cucumber-shaped element is a nonlinear element. The only thing we know about it is that it behaves like a nonlinear resistor (see voltage versus current relationship on the graph). My goal is to find out the element's \$V_X\$ (Volts) and \$I_X\$ (Amps) in this circuit. Here is why i am stuck:

I derived possible resistances Rx of the element from the graph above and then tried to apply a simple current divisor first with Rx = 1K to see if the Ix is within 1mA range and then with Rx = 2K and again check if the Ix is within 2mA range. This does not work

I don't understand whether or not i am wrong with such a logic.

edit:

We have only two possible values for RX=dV/dI. And the Thevenin of VS, RS, and RP is VTH=VS*RP/(RS+RP) and RTH=RP ∣∣ RS, with RX now in series with RTH. Plug in the two possible values for RX and you will find two boundary currents, both of which are well within one of the segments. You only have one alternative, now. The answer is captured by the boundaries. Then you get two possible ix currents ix = 0.76 when Rx = 1k (which is right) and ix = 0.6 when Rx = 2k. The reason 1k current turned out to be right is because the two boundaries both lie on the same segment. So that segment is the one that applies. It is convenient that this segment also crosses through 0.0

Answer 1

One possible way is to use a Thevenin's theorem and replace this circuit with the equivalent circuit that contains only a single RTH resistor and nonlinear element in series with the RTH resistor and Vth voltage source.

And use the load line to find the operation point.

https://www.allaboutcircuits.com/technical-articles/how-is-a-load-line-used-in-circuit-design/

The equivalent circuit will look like this:

Where:

$$V_{TH} = V_S \times \frac{R_P}{R_S + R_2} \approx 2.76V$$

$$R_{TH} =R_S||R_P \approx 2.6k\Omega$$

All this means that the current that flows through a nonlinear element cannot be greater than this: $$I = \frac{V_{TH}}{R_{TH}} \approx 1.06mA $$

Answer 2

I think that a way of answering this question is to find the Thevenin voltage and resistance for the circuit shown below and then use a load line.

Having found the Thevenin voltage, V_th, and resistance, R_th, you can add a load line for the Thevenin equivalent circuit to the V_x vs I_x graph for the non-linear element.

The load line will have an intercept of V_th on the V_x axis, a gradient of -R_th and an intercept of V_th/R_th on the I_x axis.

From that you can decide where the intercept of the two graphs, $A$ or $B$, is and hence find the values of the voltage and current at that intercept.

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Another way:
To find the position of the load line one can find the open circuit voltage across R_p (which is the Thevenin voltage) and the short circuit current, V_S/R_S,. These two values define the position of the load line.

Answer 3

You can use iteration to get the value. Start with an assumption that RC (R cucumber) is 1500 because 3V/2mA = 1500R The starting point doesn't matter much. Solve for RC and RP in parallel (RC||RP). Calculate VRPRC, the voltage across RP and RC in parallel. Then use that voltage to determine what RC would actually be at that voltage. The formaula, from the chart,

RCactual = IF(VRPRC<1,1000,2*VRPRC/(VRPRC+1)*1000)

The first iteration estimates RCactual at 1005R. So use that as the starting point for iteration 2. Iteration 2 finds that RCactual = 1000R. Use that as a starting point for iteration 3. Iteration 3 finds that RCactual =1000R. The assumption and the result agree. So RC = 1000. You now have the voltage across RC||RP and can find any voltage and current you need.